Question #276794

A crate of mass 10kg is pulled up arough incline with an initial speed of 1.5m/s. the pulling force is 100N parallel to the incline which makes an angle of 30° with the horizontal. the coefficient of kinetic friction is 0.4 and the crate is pulled 5m. How much mechanical energy is lost due to friction?

1
Expert's answer
2021-12-07T20:11:33-0500

Energy loss is simply Kinetic friction multiply by the distance traveled. To find kinetic friction, take coefficient of friction multiply by normal force acting on the block.

N=10×9.8×cos30=84.87N×k=84.87×0.4=33.95Energy  loss=33.95×5=169.74  JN = 10 \times 9.8 \times cos30 = 84.87 \\ N \times k = 84.87 \times 0.4 = 33.95 \\ Energy\; loss = 33.95 \times 5 = 169.74 \;J


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